1. 4/28 Warm-up Find the surface Area
Learning Target: To find surface area of pyramids and cones
In class notes/exit
HW: wks

2. Answers to Classwork

3. Surface Area of Pyramids, Cones, and Spheres

4. Key Definitions
Pyramid – a polyhedron in which the base is a polygon and the lateral faces are triangles with a common vertex.
Vertex of the Pyramid – The intersection of all the lateral faces.
Regular Pyramid – A regular polygon as the base and the segment joining the vertex and the center of the base is perpendicular to the base.
Slant Height – The height of the lateral face, l.

5. Vertex
Lateral
Faces
Slant Height
Base

6. A regular square pyramid has a height of 15 centimeters and a vase edge length of 16 centimeters. Find the area of each lateral face of the pyramid.
Now we can find the area of one triangle of the lateral faces
Need to find the slant height
l = 17 cm
𝐴= 1 2 𝑏ℎ
l = 17 cm
h = 15 cm
𝐴= 1 2 (16)(17)
b = 16 cm
b = 16 cm
𝐴=136 cm2
𝐿𝐴=136(4) = 544 cm2

7. A regular hexagonal pyramid and its net are shown
A regular hexagonal pyramid and its net are shown. Let s represent the length of a base edge, and let l represent the slant height of the pyramid. This will help find the surface area of the pyramid.
Lateral Area = 𝑠𝑙
Lateral Area =𝑛 1 2 𝑠𝑙
𝑛𝑠=𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒
Lateral Area = 𝑃𝑙
We represent the area of the base by… B

8. Surface Area of Regular Pyramid
𝑆=𝐵+ 1 2 𝑃𝑙
B = Area of base
P = Perimeter of base
l= Slant height

9. Key Definitions
Cone – A solid with a circular base and a vertex not in the same plane as the base
Right Cone – When the segment joining the vertex and the center of the base are perpendicular.
Lateral Surface – all segments connecting the base to the vertex.

10. When you cut along the slant height and base edge and lay it flat, the net of the cone looks like the shape below.
2πr
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑖𝑟𝑐𝑙𝑒 = 𝑀𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝐴𝑟𝑐 360
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑖𝑟𝑐𝑙𝑒 = 𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝐴𝑟𝑐 𝐿𝑒𝑛𝑔𝑡ℎ 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑀𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝐴𝑟𝑐 360
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 π 𝑙 2 = 2π𝑟 2π𝑙 l
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟= 2π𝑟 2π𝑙 ∙π 𝑙 2
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟=π𝑟𝑙 = Lateral Area

11. Surface Area of Right Cone
𝑆=𝐵+π𝑟𝑙
𝑆=π 𝑟 2 +π𝑟𝑙
B = Area of Base (circle)
r = Radius of Base
l = Slant Height

12. Example 2a l = 6 2 + 8 2 =10 𝑆=π 6 2 +π(6)(10) 𝑆=96π = 301.6 m2
𝑆=π 𝑟 2 +π𝑟𝑙
r = 6
l = =10
𝑆=π π(6)(10)
𝑆=96π = m2

13. Example 3 l = 5.7 2 + 18 2 =18.88 𝐿𝐴=π(5.7)(18.8) 𝑆=338.4 in2
𝑆=π 𝑟 2 +π𝑟𝑙
r = 5.7
l = =18.88
𝐿𝐴=π(5.7)(18.8)
𝑆=338.4 in2

14. Example 4 – you try!
LA = 375π = yd2
SA = 600π = 1884 yd2

15. WarmUp:Find the area of the circle.
4/29
WarmUp:Find the area of the circle.
Learning Target: To find surface area of spheres and composite area
In class: cont.notes/practice/extra credit due today!
HW: organize
5 in

16. Key Definitions
Sphere – the set of all points in space equidistant from a given point.
Center – The point the set of points are equidistant from.
Radius – A segment from the center point to a point on a sphere
Chord – A segment whose endpoints are on the sphere
Diameter – A chord that contains the center
Great Circle – When you intersect a plane with the sphere, the plane intersects the center of the sphere.
Hemisphere – Every great circle separates the shape into two congruent halves.

17. Picture a baseball…

18. Surface Area of a Sphere
𝑆=4π 𝑟 2

19. Example 1